%I #18 Nov 19 2018 10:33:06

%S 1,0,2,5,16,20,18,43,162,190,532,916,564,3314,3908,10499,30789,53828,

%T 153384,62946,278737,364195,629686,3768344,7827416,9496221,23159959,

%U 184328920,68035462,92566977,457932094,370110663,648634305,4032924162,7841376455

%N Least k such that prime(k) + prime(k+1) contains n prime divisors (with multiplicity), otherwise 0.

%C If p and q are two consecutive odd primes, then p + q is the product of at least three primes (not necessarily distinct) because p + q = 2*(p + q)/2 => (p + q)/2 is a composite integer between two consecutive primes p and q that is the product of at least two prime numbers. Thus 2*(p + q)/2 has at least three prime factors => a(1) = 1 because prime(1) is even => prime(1) + prime(2) = 5 is prime and a(2) = 0, probably the only 0 of the sequence.

%H Giovanni Resta, <a href="/A251600/b251600.txt">Table of n, a(n) for n = 1..45</a>

%e a(5) = 16 because prime(16) + prime(17) = 53 + 59 = 112 = 7*2^4 with 5 prime divisors.

%t A251600 = {1, 0}; Do[k = 1; While[PrimeOmega[Prime[k] + Prime[k + 1]] != n, k++]; AppendTo[A251600, k], {n, 3, 10}]; A251600

%Y Cf. A000040, A001222.

%K nonn

%O 1,3

%A _Michel Lagneau_, Dec 05 2014

%E a(28)-a(33) from _Daniel Suteu_, Nov 18 2018

%E a(34)-a(35) from _Giovanni Resta_, Nov 19 2018